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ahrayia [7]
3 years ago
8

Explain how to plot a point when given the coordinates.

Mathematics
1 answer:
velikii [3]3 years ago
7 0
Here's the answer right here

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PLS HELP !!!<br> a. 15<br> b. 1<br> c. 2<br> d. 11
Serjik [45]

Answer:

d

Step-by-step explanation:

ive had this exac problem before

5 0
3 years ago
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if p(x) = x2 – 1 and q(x)=5(x-1), which expression is equivalent to (p – q)(x)? a: 5(x – 1) – x2 – 1 b: (5x – 1) – (x2 – 1) c: (
zavuch27 [327]
P(x) = x^2 - 1
q(x) = 5(x - 1)
(p - q)(x) = (x^2 - 1) - 5(x - 1)
7 0
3 years ago
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Use the drop-down menus to complete the statements about factoring 14x2 + 6x – 7x – 3 by grouping. The GCF of the group (14x2 –
Yuliya22 [10]
14 x² + 6 x - 7 x - 3 =
= ( 14 x² - 7 x ) + ( 6 x - 3 ) =
= 7 x ( 2 x - 1 ) + 3 ( 2 x - 1 ) =
= ( 2 x - 1 ) ( 7 x + 3 )
Answer:
1.  GCF of the group ( 6 x - 3 ) is 3.
2.  The common binomial factor is 2 x - 1.
3.  The factored expression is:  ( 2 x - 1 ) ( 7 x + 3 ).
8 0
3 years ago
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PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS
Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

  • opposite side = a = 4
  • adjacent side = b = \sqrt{33}
  • hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

Then cosine which is adjacent over hypotenuse

\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

Tangent is the ratio of opposite over adjacent

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

------------------------------------------------------

Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0
1 year ago
Is 20/30 equivalent to 2/3​
kicyunya [14]

Answer:

Yes 20/30 is equivalent to 2/3

Step-by-step explanation:

Divide the fraction 20/30 by 10/10 to simply to 2/3

3 0
3 years ago
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